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Python面试宝典2020版
def more_than_half(items):
dict = {}
for item in items:
dict[item] = dict.get(item, 0) + 1
for k,v in dict.items():
if v > len(items)//2: # 题目要求多于50%
return k
题目里提到的两个问题是不等价的
找零顺序是无关的,上台阶的走法要考虑走几格的顺序
比如change_money(5) 其实只有两种方法5, 2+3
上5格台阶的话 就有三种方法,5 = 2+3 = 3+2
permutation or combination
如果列表前置位有多个元素不断重复相互覆盖次数,导致这些元素对于结果没有影响,那么后续的判断中相当于减小了列表的长度,后置位中的新元素只要出现次数多于减小后列表的50%,就会被作为最终结果输出,但是这种情况下的结果明显是不合题意的。
比如输入[0, 1, 0, 1, 2, 2, 3],输出结果为2
(没学多久,说错了请见谅😂)
感觉你的第4题的答案和解释不一致
最后打印出来的结果应该是
True True
True False
True
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